Question

Using the monochromatic light of same wavelength in the experimental set up of the diffraction pattern as well as in the interface pattern where the slit separation is 1 mm. 10 interference fringes are found to be within the central maximum of the diffraction pattern. Determine the width of the single slit if the screen is kept at the distance from the slit in the two cases.

Answer 1

Hint: We should have the basic knowledge of the phenomenon of diffraction and interference of light through a single slit. We will use the formula of total angular width. After that, we will equate both the equations of interference and diffraction. Since the wavelength is the same as the light source is the same.

Formula used:
\[\theta =\dfrac{{{\lambda }_{{}}}}{d}\]

Complete step by step answer:
Wavelength is the same D for both.
We know that If a is the width of the single slit,then for the central maximum of the single slit diffraction pattern
\[\begin{align}
  & \sin \theta \approx \theta =\dfrac{{{\lambda }_{1}}}{a} \\
 & \\
\end{align}\]
= Total angular width \[=2\theta =\dfrac{2{{\lambda }_{1}}}{a}\]
For 10 interference fringes of the double slit interference pattern to lie within the central maximum of the single split pattern.
= total angular width\[=2\theta =10\dfrac{2{{\lambda }_{2}}}{d}\]
d is the separation between the slits .
from both the equations
\[\dfrac{2{{\lambda }_{1}}}{a}=10\dfrac{2{{\lambda }_{2}}}{d}\]
Hence,
\[\begin{align}
  & \dfrac{2}{a}=\dfrac{10}{d} \\
 & \Rightarrow a=\dfrac{d}{5}=\dfrac{1}{5}mm=0.2mm \\
 & \Rightarrow a=0.2mm \\
\end{align}\]

The width of the single slit, if the screen is kept at the distance from the slit in the two cases is 0.2mm.

Additional Information:
Interference refers to the phenomenon where two waves of an equivalent kind overlap to supply a resultant wave of greater, lower, or an equivalent amplitude. Diffraction is defined because of the bending of a wave around the corners of an obstacle or aperture. Diffraction refers to varied phenomena that occur when a wave encounters an obstacle or a slit. It is defined because of the bending of waves around the corners of an obstacle or through an aperture into the region of the geometrical shadow of the obstacle/aperture.

Note:
We should note that the width of the central diffraction maximum is inversely proportional to the width of the slit. If we increase the width size a, the angle T at which the intensity first becomes zero decreases, leading to a narrower central band.